Question: Solve for $x$. Enter the solutions from least to greatest. $3x^2 - 9x - 12 = 0$ $\text{lesser }x = $
Solution: $\begin{aligned} 3x^2 - 9x - 12&= 0 \\\\ 3(x^2-3x-4)&=0 \end{aligned}$ Now let's factor the expression in the parentheses. $x^2-3x-4$ can be factored as $(x+1)(x-4)$. $\begin{aligned} 3(x+1)(x-4)&=0 \\\\ x+1=0&\text{ or }x-4=0 \\\\ x=-1&\text{ or }x=4 \end{aligned}$ In conclusion, $\begin{aligned} \text{lesser }x &= -1 \\\\ \text{greater }x &= 4 \end{aligned}$